Question #988dd

1 Answer
Apr 2, 2017

Answer:

#ZnCl_2 + Na_2CO_3 rarr ZnCO_3 + 2 NaCl#

Explanation:

#ZnCl_2 + Na_2CO_3 rarr ZnCO_3 + NaCl#

This is what I do...
I write that all elements on the reactant side as well as those on the product side. Not forgetting to split them when there is an addition.

Reactants

#Zn=1#
#Cl=2#
#+#
#Na=2#
#C=1#
#O=3#

Products

#Zn=1#
#C=1#
#O=3#
#+#
#Na=1xxcolor(red)2=2#
#Cl=1xxcolor(red)2=2#

From the above, we see that #Na# is #2# on the reactant side and #1# on the product side so we multiply by #color(red)2# to make it equal. When you do this, it automatically affects #Cl# because it forms a compound.

Check to see if all elements are balanced then rewrite the equation.

#ZnCl_2 + Na_2CO_3 rarr ZnCO_3 + color(red)2 NaCl#